Comments On The Uniform Degeneracy Of AOP

Questions and Answers

1. Why is it called uniform degeneracy?
   The degeneracy theory uses a degeneracy operator that has three unary operators in the form "^ofyw". When all the three unary operators are the same they provide a uniform action, thus I call it uniform degeneracy.
   
   The notation of the general degeneracy theory  "^offf" is used to make all materials consistent for readers when I publish "The General Degeneracy  Theory”
   
   I guess I named this operation that way due to my background in physics. In physics, two states are said to be degenerate if there is appreciable overlap of their natural widths, or if other small effects disposition their energies about by amounts comparable to their separation.  The uniform degeneracy of a prioritor, under AOP, is defined mathematically as the image of its priority assignment. Physically, what we are doing is dispositioning (repositioning) the logical values in the priority-assignment to have a new order of priorities.  This process result in a new prioritor that has a different function from the prioritor we just took its uniform degeneracy. Here we have the same parallel phenomena. Instead of "states" we have "priority-assignments". Instead of "dispositioning states" we are "dispositioning the order of priorities in the priority-assignment". Instead of "applying energy to disposition the states", we are applying "a unary image operation to disposition the priorities". Thus, the name "degeneracy" is very meaningful.
   

2. You define the uniform degeneracy under a conservative unary operator as the image of the priority-assignment of the prioritor. Why do you use "^offf" as an operator to your degeneracy not use simply the image operator "^─f" itself?
   
   The "^─f" is fine only for prioritors because each prioritor has a priority assignment of distinct digits. However, how can we take the degeneracy of the "XOR" operator in the binary system, which is not a prioritor? We can not operate on the XOR by the "^─f" operator to get its degeneracy. However, when we apply the offf degeneracy operator on the XOR or on the AND we get the right degeneracy. The paper space is so limited to provide a description to the general degeneracy theorem.
   

3. Did you type the uniform degeneracy Table?
   No I did not. I generated the table by a turbo Pascal program on a text file and then loaded it by Microsoft-Word and edited the headings. This avoids typing errors. The enclosed software package generates this table.
   

4. Explain the notation in G(X,a,C)^offf=G(X,a^offf,C^{¾ f})?

   Assume we have the following function in Boolean algebra
   G(A,B,C)= (A+B)*(A+B)+A*C+(1+C)*(B+0). In AOP we write this as G({A,B,C},{+,*},{1,0}); using sets notations we write this as G(V,a,C) where V={A,B,C},a={+,*}, C={1,0};
   
   Thus, in G(x,a,c) : 

   x: is the set of all variables used by the function
   a: is the set of all prioritors used by the function
   c: is set of all constants used by the function
   G: A function whose range is determined by a set of variables 'x' and a set of prioritors 'a' and a set of constants 'c'.

   This expression “G(x,a,C)^offf" is read as the uniform degeneracy of the function G. According to the uniform degeneracy definition, we have to take the uniform degeneracy of each prioritor and take the image of each constant and leave all variables untouched. The statement is translated symbolically as G(X,a^offf,C^─f). This means that the "^offf" uniform degeneracy is to operate on all prioritors in the set 'a' and the '^─f' image operator is to operate on all the constants of the 'C' set.

   Using the above example in Boolean algebra we have

   G(A,B,C)^offf= G({A,B,C},{+,*},{1,0})^offf
                         = G({A,B,C},{+,* } offf,{1,0}^─f)
                         = G({A,B,C},{+^offf,* ^offf},{1^─f,0^─f})  
                         =(A+^offfB)*^offf(A+^offfB)+ ^offf A*^offfC+^offf (1^─f +^offfC)*^offf(B+^offf 0^─f)

5. What is the advantage of degeneracy theoretically and practically?
   
   Theoretically: the degeneracy theory gives us the power to discover new theorems and new systems of algebras other than AOP. In terms of new theorems, assume we discovered a new equation for two prioritors. By applying the uniform degeneracy on this equation we generate z! theorems. 
   
   In terms of new algebras, AOP in the ternary system describes only 3!=6 binary operators (prioritors) out of the 19,863 operators and when we introduce the none uniform degeneracy, AOP will describe only 3! 3=216 binary operations. This number is still very small compared to the 19,863. Therefore, we conclude that there are other algebraic systems that describe the remaining operators. The discovery of these algebraic systems is much simplified by the degeneracy theory. The degeneracy theory reveals to us that the 19863 binary operators can be degenerated from 139 binary operators where each operator is called an "element" and these 139 elements generates independent (none isomorphic) algebraic systems. AOP for the ternary system, is the outcome of the study of one element out of the 139 elements. GTODE (General Theory of Digital Elements) is a comprehensive theory that deals with the study of these elements.  For example, the 16 binary operators in the binary system can be generated only from 5 elements and the 4,294,967,296 binary operators in the quaternary system can be degenerated from the 367,794 elements.
   
   Practically: The degeneracy theory allows us to use new operators to implement the same circuit.
   
   

6. Why is the number of degenerate equation is z!?
   Because every prioritor can be degenerated into z! prioritors by a^offf=a^─f. Since, the uniform-degeneracy of a prioritor is a prioritor, then we have z! degenerate equations.
   

7. You gave the clear definition of degeneracy a^offf=a^─f . What is the need for the degeneracy Table?
   
   The table gives fast results and saves the reader the process of implementing the operation each time. Specifically, when we find the uniform degeneracy of a priority equation.
   

8. Given this equation "AÙ(BÚC)=(AÙB)Ú(AÙC)" give all the 24 degenerate equations in the quaternary system where Ù=MIN and Ú=MAX? 
   After replacing Ù=MIN by Q1 and Ú=MAX by QO, we express AÙ(BÚC)=(AÙB)Ú(AÙC) as A Q1 (B QOC)=(A Q1B) QO (A Q1C)  See Table-2.
   Note that the 24 uniform degeneracy operators are obtained directly from Table-2 by picking all the prioritors in a row. For example, for the z! degeneracy forms of Q1 we pick its row in the Table-1 which reads "OIMCGA NHK6E4 LBJ582 F9D371" and the same for QO which reads "123456 789ABC DEFGHI JKLMNO".